How do I solve this problem:  Given demand curve P=400-.1Q , at what
price is total revenue maximized?

Hi!!

The first thing you must do is to find the revenue function, you can
do that simply using the revenue definition:
Revenue = quantity demanded * unit price =
        = Q * P =
        = Q * (400 - 0.1*Q) =
        = 400*Q - 0.1*Q^2

The marginal revenue (MR) is the additional revenue derived from the
sale of one additional unit, and the derivative of the revenue
function is used to determine the marginal revenue.

MR = (400*Q - 0.1*Q^2)' 


Now if revenue has a maximum it occurs when its derivative is zero, since
Marginal Revenue is the derivative of the revenue, if revenue has a
maximum it occurs when marginal revenue is zero.
So the next step is to equal the found MR funtion to zero and find
wich value of Q satisfy that. Then use this figure at the demand
function to see wich is the price that answer your question.

To check your work I will give you the answer that I found (I hope
that I did my calculations right): P = $200


I hope that this helps you. Feel free to use the clarification feature
if you need further assistance on this question or if you find
something unclear.


Regards,
livioflores-ga

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Clarification of Answer by livioflores-ga on 16 Oct 2005 16:08 PDT

Thank you for the good rating and comments.

Perfect.  Good job

99.9% correct

YOU MUST CHECK IF THE ZERO IS A MAXIMA OR MINIMA!!!!

Sorry, but you must check your zero in the marginal revenue to make
sure you are finding a maximum (maxima) of the marginal revenue
function and not the minimum (minima).  Also, make sure it is not a
relitive maxima or minima.  The marginal revinue could (although not
in this case) rise, then fall, then rise and fall again.  Thus, there
would be 3 zeros and 2 maxima and 1 minima thus you would have to
check if:
1. Whether if it was a maxima or minmia
2. To ensure that it was the ultimate (or highest) maxima.

You don't want to go to you bosses and tell them that this is the best
throughput and actuall it is the worst or there is more optimally
efficient solution.

kmclean-ga